Find the distinct left cosets of $S_{n-1}$ in the symmetric group $S_n$.
The number of elements in $S_n$ is $n!$. The number of elements in $S_{n-1}$ is $(n-1)!$. By Lagrange Theorem we have that the number of distinct left cosets of $S_{n-1}$ in $S_n$ is $n$. I have $n$ left cosets which are $S_{n-1}$, $(1 n) S_{n-1}$, $(2 n) S_{n-1}$, …, $((n-1) n)S_{n-1}$. How will I show that these $n$ cosets are distinct?
Look at the permutation $\sigma=(123...n)$. I say $ S_{n-1},\sigma S_{n-1},\sigma^2 S_{n-1},..., \sigma^{n-1} S_{n-1}$ are $n$ distinct cosets. Can you prove it? I'll give a hint. Suppose $\sigma^i S_{n-1}=\sigma^j S_{n-1}$ when $0\leq i<j<n$. What can you say about $\sigma^{j-i}$?
Hint: Observe that for any $\sigma \in S_{n-1}$, we have $\sigma(n) = n$. Now, if $\sigma \in (1~n) \cdot S_{n-1}$, then in that case what can we say about $\sigma(n)$?
